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Let's consider a block on a frictionless table. The block is connected to a fixed support on the table via a massless spring. Suppose the block is pulled aside by a distance x and then released. The potential energy of the "block+spring" system at the instant of release is taken to be 1/2 kx2 (provided potential energy is zero at x=0).

I am aware of the following.

If external forces do no work on the system and internal forces are conservative, the mechanical energy of the system remains constant.

The difference in potential energy corresponding to a conservative force is defined as negative of the work done by the force.

Now, the forces internal to the "spring+block" system are:

  1. Force by the spring on the block
  2. Force by the block on the spring

Why isn't the potential energy corresponding to the force 2 is not taken into account while computing the potential energy of the entire "system"? Is is that force 2 is nonconservative?

My question may seem wierd. But when the gravitational potential energy of a body+earth system is defined, both works (by the gravitational force of the body and by the gravitational force of the earth) are taken into account. Work due to gravitational force of body is dismissed only because of its small value.

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    $\begingroup$ You do not count the potential energy twice. It is just $1/2\cdot x^2$ in this case (after all, the support is fixed) $\endgroup$
    – Thomas
    Mar 20 at 8:25
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    $\begingroup$ But is not this the case in computation of gravitational potential energy of body+earth system? If it is there, why not here? $\endgroup$ Mar 20 at 8:39
  • $\begingroup$ It is actually always the case. See my answer below that I just posted $\endgroup$
    – Thomas
    Mar 20 at 22:04
  • $\begingroup$ "Spring potential energy" is defined for spring forces... The wall is fixed so you get the same situation as earth... You can define different potential energy for the force by the block on the spring (if it is conservative) $\endgroup$
    – PinkAura
    Mar 21 at 7:06

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Here is a force diagram of your horizontal spring-mass system and its surroundings.

enter image description here

The two sets of forces $F_{\rm s,b}$ and $F_{\rm b,s}$, and $F_{\rm s,te}$ and $F_{\rm te,s}$ are Newton third law pairs, ie equal in magnitude and opposite in direction.

Making the assumption that the mass of the table and Earth is much greater that the mass of the block we can then assume that the right hand end of the spring does not move.

In your system only the block can have kinetic energy because the spring has no mass, and only the spring can have potential energy as the block is assumed rigid and there are no other forces as a result of an interaction between the block and the spring.

Suppose the block and the left hand end of the spring move together by a small distance $\Delta x$ to the right.
The work done on the block by the spring is $F_{\rm b,s}\,\Delta x$ (which increases the blocks kinetic energy) and the work done on the spring by the block is $-F_{\rm s,b}\,\Delta x$ (which decreases the potential energy stored in the spring).
The net amount of work done by those two internal forces is zero but it has resulted in an increase in the kinetic energy of the block and a corresponding decrease in the potential energy of the spring with no change in the total mechanical energy of the spring-mass system.

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  • $\begingroup$ "The work done on the block by the spring is 𝐹b,sΔ𝑥 (which increases the blocks kinetic energy) and the work done on the spring by the block is −𝐹s,bΔ𝑥 (which decreases the potential energy stored in the spring)"... If we apply $W_{all forces} = \Delta KE$... the work done by internal forces explicitly cancel out and we get $\Delta KE$ = 0 . But clearly kinetic energy of block would be changed... Why the equation is telling lies? $\endgroup$
    – PinkAura
    Mar 20 at 12:47
  • $\begingroup$ Work done on the block by the spring = ΔKE of the system = - (Work done on the spring by the block) = - ΔPE of the system. $\endgroup$ Mar 21 at 3:26
  • $\begingroup$ @KshitizKatiyar I was thinking of applying work energy theorem for the whole system, i believe it is not related to your question... $\endgroup$
    – PinkAura
    Mar 21 at 6:58
  • $\begingroup$ @Farcher it would be great if you can point error in my comment, I've been so restless since then $\endgroup$
    – PinkAura
    Mar 21 at 7:02
  • $\begingroup$ @PinkAura The block is not doing any work on the spring. (after all it is assumned massless). The block has potential energy at the beginning and kinetic energy at the end. Its potential energy is turned into kinetic .energy. The spring does not come into it explicitly. It is implictly contained already in the spring constant $k$ in the expression for the potential energy of the block, $\endgroup$
    – Thomas
    Mar 21 at 8:22
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Your system consists of three parts: the block, the spring and the table(if the table is nailed to the floor that would be the earth). The spring basically mediates the force between the table and the block, and because it is massless the total net force applied to it must be zero and it cannot have any kinetic energy. So, in the calculation of the mathematical construct that we call the potential energy of a massless spring, $\int F_{\text{net on spring}}.dx$, it won't have a contribution.

Your analogy actually holds here too, however, instead of the spring the other object would be the table (or the earth) which is assumed to be much heavier than the block.

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    $\begingroup$ I'm afraid I chose "block+spring" as the system. The choice of the system is independent of anything else. $\endgroup$ Mar 20 at 8:04
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    $\begingroup$ So what is the net force on the spring? @KshitizKatiyar $\endgroup$
    – Ali
    Mar 20 at 8:56
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    $\begingroup$ The only external force on the spring is that by the wall. $\endgroup$ Mar 20 at 9:52
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You do not count the potential energy twice. It is always between a pair of particles/objects. How a change of the potential energy is distributed between the two bodies as a result of the work done depends on their masses. If you have one mass a lot larger, then the kinetic energy will practically only change for the smaller mass. For instance, if you drop an object with mass $m$ in the gravitational field of the Earth by a height $h$, this corresponds to a decrease of potential energy (and an increase of kinetic energy) of $\Delta U =mgh$, not $\Delta U =2mgh$ (where $g$ is the gravitational acceleration). The Earth will practically not gain any kinetic energy at all. Similarly, in your case (with the spring support fixed) the block will increase its kinetic energy by $\Delta U =1/2\cdot kx^2$ not by $\Delta U =kx^2$. If you have two equal freely moving masses instead, they will share the potential energy between them, so each will only gain half the kinetic energy. It is always only the total kinetic energy change that is equal to the potential energy change.

I have illustrated this in the graphics below. The top half shows the case for the spring fixed to a wall ( or a very heavy free mass), the bottom half for the spring connecting two equal masses. In both cases the spring is stretched to length $x$ initially (for simplicity assuming the spring has zero extension when relaxed) with the masses at rest. The second picture in each case shows the spring contracted back to length $x/2$ with the masses having gained the corresponding kinetic energy, with the total energy (potential energy $U$ + kinetic energy $K$) equal to the initial (potential) energy $1/2\cdot k\cdot x^2$ (which is solely the elastic energy stored in the spring when stretched and is independent of the attached masses). The difference between the two cases here is merely that the kinetic energy imparted after contraction of the spring is shared between the two masses in the second case, because the work done on each mass is smaller due to the smaller distance each one moved.

potential and kinetic energy of spring system

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  • $\begingroup$ I completely understood what you meant by "not counting twice". Note my comment on the response by @Farcher . $\endgroup$ Mar 21 at 3:29
  • $\begingroup$ @KshitizKatiyar I'll add some illustrating graphics later today, which should help you to understand it better. $\endgroup$
    – Thomas
    Mar 21 at 8:24
  • $\begingroup$ @KshitizKatiyar Please see my extended answer $\endgroup$
    – Thomas
    Mar 21 at 21:25
  • $\begingroup$ If you have two equal freely moving masses instead,they will share the potential energy between them I beg to differ as there is no gravitational potential energy involved. The two masses that you mention do not store the potential energy pe as they do not change shape, form etc.The pe is stored by the spring which can be thought of as being made up of microscopic "masses" and bonds (springs) and the two masses at the ends of the spring exert forces on the spring which change the pe stored in the spring.The masses at the end are the parts of the system which has kinetic energy. $\endgroup$
    – Farcher
    Mar 22 at 11:28
  • $\begingroup$ @Farcher The potential energy of the system depends solely on the extension of the spring. After the spring has contracted by some amount, the corresponding change in potential energy will be turned into kinetic energy of the attached masses (as the spring does work on them) according to the inverse ratio of their masses, so equally for identical masses. $\endgroup$
    – Thomas
    Mar 22 at 18:26
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Since the spring is massless, the force by the block on spring does work which increases the potential energy due to change in the length of spring. The total potential energy of the system is the potential energy gained by the block when stretched or released which later gets converted to kinetic energy of block and spring. Hence, spring is massless, the kinetic energy of spring is zero which means the kinetic energy of the block.

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    $\begingroup$ Could you be clear? $\endgroup$ Mar 20 at 7:56
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    $\begingroup$ how much more clarity do you need ? i explained it in simple terms ,where you did not understand it. $\endgroup$
    – nagastar
    Mar 20 at 8:01
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    $\begingroup$ workdone due to force by the block on spring is taken as the increased potential energy of the block(also the spring's) ,which later converts to kinetic energy of the block+(spring=0) $\endgroup$
    – nagastar
    Mar 20 at 8:04
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    $\begingroup$ Potential energy of the system is due to the individual contributions of potential energies corresponding to the internal forces of the system. Why is it that you just consider the work done by the force of the block on the spring (and compute potential energy) and not the work done by the force of the spring on block? $\endgroup$ Mar 20 at 8:14
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    $\begingroup$ workdone by the force of the block on the string is while we strech it which is the gain in potential energy of the spring+block which later is converted to gain in kinetic energy of the spring+block, $\endgroup$
    – nagastar
    Mar 20 at 8:25
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We must take account of both forces...

Once released...

The force of the spring on the block, acting through a displacement in the same direction as the force, gives the block kinetic energy.

The force of the block on the spring, acting in the opposite direction to the displacement of the end of the spring takes elastic potential energy away from the spring.

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